題目概要：
計算出每個人平均要付多少錢。
解題方向：
每個case的第一個數字為總人數(N)，第二個數字為多少人平分(F)，接下來的數字為每個人所應該要付的錢(啤酒單價)。啤酒單價會超過Lnog的範圍，故需要用到Java的api(BigInteger)來實作。如過不使用BigInteger的方法，那就只能字元判斷加總為多少or相除為多少(可參考C++程式碼);
程式碼：
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| //Java | |
| import java.util.Scanner; | |
| import java.math.BigInteger; | |
| class uva10925{ | |
| public static void main(String args[]){ | |
| Scanner sc=new Scanner(System.in); | |
| int count=1; | |
| int N=0,F=0; //N=總人數，F=多少人平分價錢 | |
| while(((N=sc.nextInt())!=0 && (F=sc.nextInt())!=0)&& sc.hasNext()){ | |
| //初始化 sum=0(因為會超過Long範圍，所以使用Java所提供的api) | |
| BigInteger sum=BigInteger.ZERO; | |
| //加總 | |
| for(int i=0;i<N;i++){ | |
| BigInteger temp=sc.nextBigInteger(); | |
| sum=sum.add(temp); | |
| } | |
| //輸出資料 | |
| System.out.println("Bill #"+count+" costs "+sum+": each friend should pay "+(sum.divide(new BigInteger(Integer.toString(F))))); | |
| //Uva online Judge不能直接在上面的輸出最後面加上\r\n，會Presentation Error (PE) | |
| System.out.println(""); | |
| count++; | |
| } | |
| } | |
| } |
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| //C++ | |
| #include <iostream> | |
| #include <algorithm> | |
| #include <string.h> | |
| using namespace std; | |
| int find(char[], int); | |
| int main() { | |
| int N = 0; int F = 0; //N=總人數，F=平分人數。 | |
| int count = 1; | |
| while ((cin >> N >> F) && (N && F)) { | |
| char sum[25]; | |
| //初始化 sum陣列為0 | |
| for (int i = 0; i<sizeof(sum); i++) { | |
| sum[i] = '0'; | |
| } | |
| //add | |
| for (int i = 0; i < N; i++) { | |
| int c = 0; //進位數字 | |
| char st[25] = { '0' }; | |
| cin >> st; | |
| reverse(st, st + strlen(st)); //把剛剛所輸入的數字全部反轉，這樣方便加總。 | |
| for (int j = 0; j < sizeof(sum); j++) { | |
| if (j < strlen(st)) { | |
| int temp = sum[j] - 48 + st[j] - 48 + c; | |
| c = temp / 10; | |
| sum[j] = temp % 10 + 48; | |
| }else { | |
| int temp = sum[j] - 48 + c; | |
| c = temp / 10; | |
| sum[j] = temp % 10 + 48; | |
| } | |
| } | |
| } | |
| //division | |
| reverse(sum, sum + sizeof(sum)); //把剛剛加總的數字再反轉回來。 | |
| char di[25]; | |
| for (int i = 0; i < sizeof(di); i++) { | |
| di[i] = '0'; | |
| } | |
| int R = 0; | |
| for (int i = 1; i < sizeof(sum); i++) { | |
| di[i] = ((R * 10 + sum[i] - 48) / F) + 48; | |
| R = (R * 10 + sum[i] - 48) % F; | |
| } | |
| //Output | |
| cout << "Bill #" << count << " costs "; | |
| //find函數為找出第一個非數字的index。 | |
| for (int i = find(sum, sizeof(sum)); i<sizeof(sum); i++) { | |
| cout << sum[i]; | |
| } | |
| cout << ": each friend should pay "; | |
| for (int i = find(di, sizeof(di)); i<sizeof(sum); i++) { | |
| cout << di[i]; | |
| } | |
| cout << endl << endl; | |
| count++; | |
| } | |
| return 0; | |
| } | |
| int find(char f[], int size) { | |
| for (int i = 0; i < size; i++) { | |
| if (f[i] != '0') { | |
| return i; | |
| } | |
| } | |
| return size - 1; | |
| } |
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